Magnitude of Magnetic Field - General Physics - Solved Past Paper, Exams of Physics

Download Magnitude of Magnetic Field - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 4. (20 pts) A solenoid has 68 turns of wire wrapped around a cylindrical form 12 mm in diameter and 8 cm long. There is a current of 0.77 A flowing through the solenoid. a) (5 pts) Work out the inductance of this solenoid. L = µ0N 2A ℓ = (4π × 10−7)(68)2(π)(6 × 10−3)2 0.08 = 8.2 µH b) (5 pts) Find the magnitude of the magnetic field within the solenoid. B = µ0nI = µ0 ( N ℓ ) I = (4π × 10−7)(68)(0.77) 0.08 = 822 µT c) (5 pts) Compute the energy density in this magnetic field. uB = 1 2 B2 µ0 = ( 1 2 ) ( (822 × 10−6)2 4π × 10−7 ) = 269 mJ d) (5 pts) Calculate the total energy contained in the magnetic field within the solenoid. There are two approaches. First, you could use the energy density from (c) and the volumen of the solenoid: U = uBV = (269 × 10 −3)(π)(6 × 10−3)2(0.08) = 2.4 µJ Second, you could just use the expression for energy stored in an inductor, U = 1 2 LI2 = (8.2 × 10−6)(0.77)2 2 = 2.4 µJ